How to make random.choice in a for loop give the same output every time

Question

I am reusing a code written by someone else. It has the following line included:

 df = (random.choice(x) for i in repeat(0))

However, because there is no random.seed, the output is different every time I rerun it, which makes sense.

I tried to fix this with random.Random(500).choice(x), but since it's a for loop, it will then give me the same value for every iteration, which is not what I want.

Does anyone know how I make this pseudo-random? Thanks!

What you want is not clear. Do you want n random values from a pool? — mozway, Sep 20 '21 at 11:49
If you want the same random value from an iterable `x` repeated `n` times: `[random.choice(x)]*n` — mozway, Sep 20 '21 at 12:13
Just assign the ``random.Random(500)`` outside of the loop and use it inside? — MisterMiyagi, Sep 20 '21 at 12:19

no comment · Answer 1 · 2021-09-20T12:26:06.420

0

If the variable x doesn't get reassigned:

df = map(random.Random(500).choice, repeat(x))

If it does:

df = (choice(x)
      for choice in [random.Random(500).choice]
      for _ in repeat(0))

Try it online! (see the "Output" section)

edited Sep 20 '21 at 12:26

answered Sep 20 '21 at 12:15

no comment

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score -1 · Answer 2 · edited Sep 20 '21 at 12:04

-1

import random

nums = [1, 2, 3, 4, 5]

def repeat_random(choices):
    i = 0
    chosen = random.choice(choices)
    while i < 5:
        print(chosen)
        i += 1

repeat_random(nums)

# This will print a random number from the nums list 5 times

edited Sep 20 '21 at 12:04

Ghost Ops

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answered Sep 20 '21 at 11:54

5idneyD

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How to make random.choice in a for loop give the same output every time

2 Answers2